Question

The mass of bicycle rider along with the bicycle is 100 kg. He wants to cross over a circular turn of radius 100 m with a speed of 10 $m{s}^{-1}$. If the coefficient of friction between the tyres and the road is 0.6, the frictional force required by the rider to cross the turn, is

• A. 300 N
• B. 600 N
• C. 1200 N
• D. 150 N

Answer 1

Answer: (B)

600 N

The centripetal force acting on the bicycle is given by,

$F_c=\frac{m{v}^2}{r}$

$=\frac{100\times 10\times 10}{100}=100N$

Required frictional force to cross the turn,
$F_f=\mu mg$

$=0.6\times 100\times 10=600N$

As the frictional force is greater than the centripetal force, so the rider will be able to cross the turn.